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The question asks HOW TO DESIGN, you don't have to write the whole DFA, but you can if you want to. Just describe how you would design. There are 32 strings of length 5. Some of them do not contain a …

Consider an alphabet $\Sigma = \{0,1\}$ and the set of all infinite length strings over $\Sigma$. Then this set is of course infinite, but uncountable which can be easily proved by the diagonalization …

...but I don't understand why the $\phi_{cell}$ formula is required. The tableau for $N$ in the Sipser's proof represents a single branch of the computation of $N$. The formula $\phi_{cell} \wedg …

Think over how to convert binary representation of $n$ into unary one using two tapes. The result should be stored on the second tape. For example, if $n=6$ then the unary representation could be $111 …

$L^2 = LL = \{uv | u,v\in L\}$. In words, $L^2$ is a set of all strings that formed by concatenation of all strings from $L$. For instance, if $u=a^kb^k \in L$ and $v=a^tb^t \in L$ then $uv = a^kb^ka^ …

Note that this language excludes all strings of length divisible by 3 but not by 2, i.e., of length $3, 9, 15, 21,\dots$. Consider languages $$L_1 = \{w\mid |w|=2k, k\geq 0\}$$ and $$L_2 = \{w\mid …

$k$ means that if $w \in R^{k}_{ij}$ then on reading $w$, FA starting from the state $i$ enters the state $j$ and never enters a state $t$ where $t > k$ while reading input $w$. In the book, you learn …

The language $$L= (0+1)^*$$ (the set of all strings over $0$ and $1$) is countable. Furthermore, any subset of $L$ is also countable. However, the set of all sublanguages of $L$ $$S = P(L) = \{M \mid …

Try the following grammar (which can be transformed into a regular grammar): $S \rightarrow A \mid E \mid I \mid O\mid U$ $A \rightarrow aA \mid bA \mid cA \mid \dots \mid zA \mid \epsilon$ (all con …

Take $r = (r_1 + r_2)^*$. We have to prove $$(r_1 + r_2)^* = r_1(r_1 + r_2)^* + r_2$$ First note that both $(r_1 + r_2)^*$ and $r_1(r_1 + r_2)^* + r_2$ contain $\epsilon$ since we are given that $\e …

Use De Morgan rule $A\cup B = \overline{\overline{A} \cap \overline{B}}$. Also you could prove it using DFA for $L_1$ and $L_2$ without using complement and intersection. The latter is systematic an …

The Rice's theorem can also be stated in terms of index sets of TMs. Lets start with a basic definition of a property of a language. A property of a language is a set of languages. For example $$P_{ …

The empty language is recursive. The TM accepting this language just returns $0$/REJECT on any input. Similarly, the language $\Sigma^*$ is also recursive. Just return $1$/ACCEPT on any input. Basi …

Given a regular expression constructing an equivalent NFA with epsilon is easier than constructing equivalent DFA. Also given two DFAs you can easily construct NFAs with epsilon moves accept concaten …

In fact you are given two languages $L_1$ defined as a set of strings of balanced parentheses. $L_2$ defined as a set of strings with equal number of ('s and )'s and every prefix of w contains at …